Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $t \neq 0$. $a = \dfrac{t^2 - 6t - 40}{-8t - 32} \times \dfrac{-7t - 35}{t^2 - 10t} $
First factor the quadratic. $a = \dfrac{(t - 10)(t + 4)}{-8t - 32} \times \dfrac{-7t - 35}{t^2 - 10t} $ Then factor out any other terms. $a = \dfrac{(t - 10)(t + 4)}{-8(t + 4)} \times \dfrac{-7(t + 5)}{t(t - 10)} $ Then multiply the two numerators and multiply the two denominators. $a = \dfrac{ (t - 10)(t + 4) \times -7(t + 5) } { -8(t + 4) \times t(t - 10) } $ $a = \dfrac{ -7(t - 10)(t + 4)(t + 5)}{ -8t(t + 4)(t - 10)} $ Notice that $(t + 4)$ and $(t - 10)$ appear in both the numerator and denominator so we can cancel them. $a = \dfrac{ -7\cancel{(t - 10)}(t + 4)(t + 5)}{ -8t(t + 4)\cancel{(t - 10)}} $ We are dividing by $t - 10$ , so $t - 10 \neq 0$ Therefore, $t \neq 10$ $a = \dfrac{ -7\cancel{(t - 10)}\cancel{(t + 4)}(t + 5)}{ -8t\cancel{(t + 4)}\cancel{(t - 10)}} $ We are dividing by $t + 4$ , so $t + 4 \neq 0$ Therefore, $t \neq -4$ $a = \dfrac{-7(t + 5)}{-8t} $ $a = \dfrac{7(t + 5)}{8t} ; \space t \neq 10 ; \space t \neq -4 $